91Ó°ÊÓ

1. Refractive index of flat-glass disk,$n=1.5$
2. Radius of curvature of the machine,$R=40\text{cm or}60\text{cm}$

A lens is a combination of two mirrors and thus has two radii of curvature. For this given system of the lens with two radii of curvature and two foci, we use the lens-maker equation to calculate the combined focal length of the given lens and the required data.

Formula:

The focal length of the lens-maker formula,

$\frac{\mathbf{1}}{\mathbf{f}}\mathbf{=}\mathbf{(}\mathbf{n}\mathbf{-}\mathbf{1}\mathbf{)}\mathbf{(}\frac{\mathbf{1}}{{\mathbf{R}}_{\mathbf{1}}}\mathbf{-}\frac{\mathbf{1}}{{\mathbf{R}}_{\mathbf{2}}}\mathbf{)}$ ...(i)

(a)

For biconvex lens,$R_1=40\text{cm,}{R}_2=-40\text{cm}$

Thus, the focal length for lens 1 can be calculated using equation (i) as follows:

role="math" localid="1663027919770" $$\begin{gathered} \frac{1}{f}=\left(1.5-1\right)\left(\frac{1}{40\text{cm}}-\frac{1}{-40\text{cm}}\right) \\ =\left(0.5\right)\left(\frac{-40-40}{40×\left(-40\right)\text{cm}}\right) \\ =\left(0.5\right)\left(\frac{-80}{-1600\text{cm}}\right) \\ =\frac{1}{40\text{cm}} \\ f=40\text{cm} \end{gathered}$$

Hence, the value of focal length is $40\text{cm}$.

(b)

Since $f>0$, the lens forms a real image of the Sun.

Hence, the image is real.

(c)

For planar convex lens,$R_1=∞\text{,}{R}_2=-40\text{cm}$

Thus, the focal length for lens 2 can be calculated using equation (i) as follows:

$$\begin{gathered} \frac{1}{f}=\left(1.5-1\right)\left(\frac{1}{∞}-\frac{1}{-40\text{cm}}\right) \\ =\left(0.5\right)\left(0-\frac{1}{-40\text{cm}}\right) \\ =\left(0.5\right)\left(\frac{1}{40\text{cm}}\right) \\ =\frac{1}{80\text{cm}} \\ f=80\text{cm} \end{gathered}$$

Hence, the value of focal length is $80\text{cm}$.

(d)

Since $f>0$, the lens forms a real image of the Sun.

Hence, the image is real.

(e)

For meniscus convex lens,$R_1=40\text{cm,}{R}_2=60\text{cm}$

Thus, the focal length for lens 3 can be calculated using equation (i) as follows:

role="math" localid="1663028569861" $$\begin{gathered} \frac{1}{f}=\left(1.5-1\right)\left(\frac{1}{40\text{cm}}-\frac{1}{60\text{cm}}\right) \\ =\left(0.5\right)\left(\frac{60\text{cm}-40\text{cm}}{(60\text{cm})(40\text{cm})}\right) \\ =\left(0.5\right)\left(\frac{20}{2400\text{cm}}\right) \\ =\left(0.5\right)\left(\frac{1}{120\text{cm}}\right) \\ =\frac{{\displaystyle 1}}{{\displaystyle 240\text{cm}}} \\ f=240\text{cm or}2.4\text{m} \end{gathered}$$

Hence, the value of focal length is $240\text{cm or}2.4\text{m}$.

(f)

Since $f>0$, the lens forms a real image of the Sun.

Hence, the image is real.

(g)

For biconcave lens,$R_1=-40\text{cm,}{R}_2=40\text{cm}$

Thus, the focal length for lens 4 can be calculated using equation (i) as follows:

role="math" localid="1663029162925" $$\begin{gathered} \frac{1}{f}=\left(1.5-1\right)\left(\frac{1}{-40\text{cm}}-\frac{1}{\text{40cm}}\right) \\ =\left(0.5\right)\left(\frac{40\text{cm}+40\text{cm}}{(-40\text{cm})(40\text{cm})}\right) \\ =\left(0.5\right)\left(\frac{-80}{1600\text{cm}}\right) \\ =\left(0.5\right)\left(\frac{-1}{20\text{cm}}\right) \\ =\frac{-1}{4\text{cm}} \\ f=-4\text{cm} \end{gathered}$$

Hence, the value of focal length is$-4\text{cm}$.

(h)

Since $f<0$, the lens forms a virtual image of the Sun.

Hence, the image is virtual.

(i)

For planar concave lens, $R_1=∞\text{,}{R}_2=40\text{cm}$

Thus, the focal length for lens 4 can be calculated using equation (i) as follows:

$$\begin{gathered} \frac{1}{f}=\left(1.5-1\right)\left(\frac{1}{∞}-\frac{1}{\text{40}\text{cm}}\right) \\ =\left(0.5\right)\left(0-\frac{1}{\text{40}\text{cm}}\right) \\ =\left(0.5\right)\left(\frac{-1}{40\text{cm}}\right) \\ =\frac{-1}{\text{80}\text{cm}} \\ f=-80\text{cm} \end{gathered}$$

Hence, the value of focal length is role="math" localid="1663029521117" $-80\text{cm}$.

(j)

Since $f<0$, the lens forms a virtual image of the Sun.

Hence, the image is virtual

(k)

For meniscus concave lens,$R_1=60\text{cm,}{R}_2=40\text{cm}$

Thus, the focal length for lens 6 can be calculated using equation (i) as follows:

$$\begin{gathered} \frac{1}{f}=\left(1.5-1\right)\left(\frac{1}{\text{60}\text{cm}}-\frac{1}{\text{40}\text{cm}}\right) \\ =\left(0.5\right)\left(\frac{40\text{cm}-60\text{cm}}{(60\text{cm})(40\text{cm})}\right) \\ =\left(0.5\right)\left(\frac{-20}{2400\text{cm}}\right) \\ =\left(0.5\right)\left(\frac{-1}{120\text{cm}}\right) \\ =\frac{-1}{\text{240}\text{cm}} \\ f=-240\text{cm or}-\text{2.4}\text{m} \end{gathered}$$

Hence, the value of focal length is $-240\text{cm or}-\text{2.4}\text{m}$.

(l)

Since $f<0$, the lens forms a virtual image of the Sun.

Hence, the image is virtual.